C = \(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{12+2\sqrt{\left(\sqrt{13}+1\right)^2}}-\sqrt{\left(\sqrt{11}+1\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{14+2\sqrt{13}}-\left(\sqrt{11}+1\right)\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

C = \(\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{13}+\sqrt{11}\right)\)

C \(\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)\) = \(13-11\) = \(2\)

\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)

\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)

sao dấu= thứ 2 lại ra như vậy

sai ở bc 2 rồi

\(H=2\sqrt{27}+\sqrt{243}-6\sqrt{12}\\ =2\cdot\sqrt{9}\cdot\sqrt{3}+\sqrt{81}\cdot\sqrt{3}-6\cdot\sqrt{4}\cdot\sqrt{3}\\ =2\cdot3\cdot\sqrt{3}+9\cdot\sqrt{3}-6\cdot2\cdot\sqrt{3}\\ =6\sqrt{3}+9\sqrt{3}-12\sqrt{3}\\ =3\sqrt{3}=\sqrt{9}\cdot\sqrt{3}=\sqrt{27}\)

\(I=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\\ =\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\\ =\sqrt{\sqrt{13}^2-2\cdot\sqrt{13}\cdot1+1^2}+\sqrt{\sqrt{13}^2+2\cdot\sqrt{13}\cdot1+1^2}\\ =\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\\ =\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\\ =\sqrt{13}-1+\sqrt{13}+1\\ =2\sqrt{13}=\sqrt{4}\cdot\sqrt{13}=\sqrt{52}\)

\(I=\sqrt{10-4\sqrt{6}}+\sqrt{10+4\sqrt{6}}\\ =\sqrt{6-2\cdot\sqrt{6}\cdot2+4}+\sqrt{6+2\cdot\sqrt{6}\cdot2+4}\\ =\sqrt{\sqrt{6}^2-2\cdot\sqrt{6}\cdot2+2^2}+\sqrt{\sqrt{6}^2+2\cdot\sqrt{6}\cdot2+2^2}\\ =\sqrt{\left(\sqrt{6}-2\right)^2}+\sqrt{\left(\sqrt{6}+2\right)^2}\\ =\left|\sqrt{6}-2\right|+\left|\sqrt{6}+2\right|\\ =\sqrt{6}-2+\sqrt{6}+2\\ =2\sqrt{6}=\sqrt{4}\cdot\sqrt{6}=\sqrt{24}\)

\(H=2\sqrt{27}+\sqrt{243}-6\sqrt{12}=6\sqrt{3}+9\sqrt{3}-12\sqrt{3}=3\sqrt{3}\)

\(I=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)

\(I=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)

K đề đọc o hiểu

L đề đọc o hiểu

Với n > 0 Ta có:

\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)

\(=\sqrt{n+1}+\sqrt{n}\)

\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)

\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)

\(\sqrt{16}+\sqrt{9}=3+4=7\)

a) \(3\sqrt{2}-2\sqrt{3}=\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\)

b) \(\sqrt{2}+\sqrt{6}+\sqrt{14}+\sqrt{42}=\sqrt{2}\left(1+\sqrt{3}+\sqrt{7}+\sqrt{21}\right)\)

\(=\sqrt{2}\left(1+\sqrt{3}\right)\left(1+\sqrt{7}\right)\)

c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}=\dfrac{\sqrt{6}}{2}\)

a) \(3\sqrt{2}-2\sqrt{3}=\sqrt{3}.\sqrt{3}.\sqrt{2}-\sqrt{2}.\sqrt{2}.\sqrt{3}=\left(\sqrt{3}-\sqrt{2}\right).\sqrt{6}\)

b) \(\sqrt{2}+\sqrt{6}+\sqrt{14}+\sqrt{42}=\left(\sqrt{3}+1\right)\sqrt{2}+\sqrt{14}\left(\sqrt{3}+1\right)=\sqrt{2}\left(\sqrt{7}+1\right)\left(\sqrt{3}+1\right)\)

c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{2}}=\sqrt{\dfrac{9}{4}}\)

\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)

\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)

\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)

\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)

\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ

\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)

\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)

\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)

\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)

#Học tốt ạ

a. \(VT=\sqrt{14+2\sqrt{13}}-\sqrt{14-2\sqrt{13}}\)

=\(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{\left(\sqrt{13}-1\right)^2}=\sqrt{13}+1-\left(\sqrt{13}-1\right)\)

\(=\sqrt{13}+1-\sqrt{13}+1=2=VP\left(đpcm\right)\)

b. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{5-2\sqrt{6}}-\sqrt{2}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{2}\)

\(=2+\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{2}=2+\sqrt{3}-\sqrt{3}+\sqrt{2}-\sqrt{2}\)

\(=2=VP\left(đpcm\right)\)

Áp dụng HĐT số 3 ta có :

 \(B=\sqrt{14-2\sqrt{3}}+\sqrt{14+2\sqrt{3}}\)

\(=\left(\sqrt{14}\right)^2-\left(2\sqrt{3}\right)^2\)